The following example illustrates how both principles of probability can be comb
ID: 197201 • Letter: T
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The following example illustrates how both principles of probability can be combined to solve problems in genetics. Probability can also be a useful tool in predicting the results of dihybrid and trihybrid matings. For example, in mating AaBb × AaBb, one would expect 1/4 of the progeny to be AA, 2/4 Aa, and 1/4 aa, or 3/4 4- to 1/4 aa. Likewise, 1/4 should be BB, 2/4 Bb, and 1/4 bb, or 3/4 B- to 1/4 bb. Now, if the A(a) and B(b) genes are independently assorting, one would expect 34 × 3/4 = 9/ proach in answering the following series of questions. If AaBb probability that the offspring will have 16 of the progeny to be A-B-,1/4 × 1/4 to be AAbb, and so forth. Use this ap- is mated to AaBb, what is the /y) a. either the phenotype aal-or the genotype aabb Lyyx31.) b, either the genotype aabb or the genotype AaBb? 4-x-y aabb or the genotype Aabbyx -nara) t (½ 2): 5/16 1 Note that the theoretical expectation from the cross Cex cc is a 3:1 ratio. However, when only our progeny are pro- duced, the probability is actually less than one balf (27/64) that the expected 3:1 ratio will be obtained! Thus, geneticists generally favor experimental organisms that produce many progeny so that theoretically expected ratios can be more read ily approximatedExplanation / Answer
Applying the Mendelian Laws of Inheritance (Segregation, Independent Assortment, Dominance):
Genotype = AaBb * AaBb
Gametes = AB, Ab, aB, ab * AB, Ab, aB, ab
Possibility of getting Phenotype aaB- = 1/4 chances of aa * 3/4 chances of BB, Bb, bB = 3/16
Possibility of getting Genotype aabb = 1/4 chances of aa * 1/4 chances of bb = 1/16
Possibility of getting Phenotype aaB- or Genotype aabb = 3/16 + 1/16 = 1/4
Possibility of getting Genotype aabb = 1/4 chances of aa * 1/4 chances of bb = 1/16
Possibility of getting Genotype AaBb = 1/2 chances of aa * 1/2 chances of bb = 1/4 = 4/16
Possibility of getting Genotype aabb or Genotype AaBb = 1/16 + 4/16 = 5/16
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